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If a Weil divisor $D$ on a normal variety is not Cartier at $x$, every linearly equivalent divisor has $x$ in its support of a Weil divisor. Otherwise that representative would vanish near $x$, expressing $D$ as a principal Weil divisor there. On $V(ac-b^2)$ the prime divisor $V(a,b)$ is not Cartier at the origin: its ideal has two independent generators modulo the maximal ideal times that ideal. This gives an unavoidable support point even for representatives with negative coefficients.